# Question on Compact operators: Suitable definition of a sequence of compact operators

Let $$H$$ denote a separable Hilbert space with an orthonormal basis $$\{e_k\}_k\in \mathbb N$$ and consider a linear, bounded operator $$A:H \to H$$ such that: $$Ae_k=\lambda_k e_k$$. Show that $$T$$ is a compact operator provided that $$\lambda_k \to 0$$ as $$k\to \infty$$.

Following the argument which can be found here or in other similar posts, I want to define a sequence of operators $$A_n$$ which will be of finite Rank (thus compact) and then using the fact that $$\lambda_k \to 0$$, to deduce that $$A_n$$ converges to $$A$$ and hence $$A$$ is also compact.

My question is: How should I define $$A_n$$?

Unfortunately I can't think anything that fits here. The fact that I have $$A$$ mapping $$e_k$$ and not a general $$x \in l^2$$ confused me a lot.

Any help or hint is much appreciated. Thanks in advance!

Any element $$x$$ of $$H$$ has an expansion $$x = \sum\limits_{k=1}^{\infty} a_i e_i$$. Define $$A_n(\sum\limits_{k=1}^{\infty} a_i e_i)= \sum\limits_{k=1}^{n} a_i\lambda_ie_i$$. The range of $$A_n$$ is contained is the span of $$\{e_1,e_2,...,e_n\}$$. I leave it to you to verify that $$\|A-A_n\| \leq \sup \{|\lambda_i|: i>n\} \to 0$$.
• $\|(A-A_n)(\sum a_ie_i)\|^{2} = \sum\limits_{k=n+1}^{\infty} |\lambda_i a_ie_i|^{2} \leq sup_{i>n} |\lambda_i|^{2} \|x\|^{2}$ if $x=\sum a_ie_i$. – Kavi Rama Murthy Jan 30 '19 at 12:20
• Yes sorry I didn't read that $e_n$ is orthonormal basis. But then indeed it's true what you have written there. Sorry again for the misunderstanding. (+1) – Shashi Jan 30 '19 at 12:21